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Maybe too easy a question for most members of this site, but suppose whenever $F$ and $G$ belong to the Selberg class, then so does $F\otimes G$ where the considered tensor product of $F$ and $G$ is defined in this paper. Does this imply that the degree of $F\otimes G$ is the product of the degrees of $F$ and $G$?

Thanks in advance.

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So far as I know, the "Rankin-Selberg" tensor product should be the same thing in the Selberg-class world as in the Langlands-conjecture world. Certainly all known cases are consistent, to my knowledge.

If we believe that, then the $p$th Euler factor for the Rankin-Selberg product would be $1/\det(1-p^{-s}A_p\otimes B_p)$, where the $p$th Euler factors for the two individuals are $1/\det(1-p^{-s}A_p)$ and $1/\det(1-p^{-s}B_p)$.

So the degree of the tensor product $L$-function would be the product of the degrees of the individuals.

(The subordinate question of the correctness of the cited paper's sense of "tensor product" is separate...)

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